Chapter 7, Problem 7.10EP

For the circuit in Figure 7.41(a), the parameters are, $R_1=200\text{kΩ}$ , $R_2=220\text{kΩ}$ $R_C=2.2\text{kΩ}$ , $R_C=2.2\text{kΩ}$ , $R_L=4.7\text{kΩ}$ , $R_E=1\text{kΩ}$ , $r_s=100\text{kΩ}$ , and $V_{CC}=5\text{V}$ . The transistor parameters are ${\beta }_o=100$ , $V_{BE}\text{(on)=0}\text{.7V}$ , $V_A=\infty$ , and $C_{\pi }=1\text{pF}$ . Consider the simplified hybrid− $\pi$ model of the transistor. (a) Determine the midband current gain $A_i=I_o/I_i$. (b) Find the Miller capacitance $C_M$ for (i) $C_{\mu }=0$ and (ii) $C_{\mu }=0.08\text{pF}$ . (c) Determine the upper 3dB frequency for
(i) $C_{\mu }=0$ and (ii) $C_{\mu }=0.08\text{pF}$ . (Ans. (a) $A_i=-30.24$ ; (b) (i) $C_M=0$ , (ii) $C_M=4.38\text{pF}$ ; (c) $f_{3dB}=60.2\text{MHz}$ , (ii) $f_{3dB}=11.2\text{MHz}$ )

To determine

The midband current gain.

Answer to Problem 7.10EP

The value of mid-band current gain $A_i$ is -30.24A/A.

Explanation of Solution

Given:

The parameters for the given circuit:

  $\begin{array}{l}{R}_1=200\text{k}\Omega \\ R_2=220\text{k}\Omega \\ R_C=2.2\text{k}\Omega \\ R_L=4.7\text{k}\Omega \\ R_E=1\text{k}\Omega \\ r_S=100\text{k}\Omega \\ V_{CC}=5\text{V}\\ {\beta }_o=100\\ V_{BE\left(on\right)}=0.7\text{V}\\ V_A=\infty \\ C_{\pi }=1\text{pF}\end{array}$

Drawing the DC equivalent circuit for the given circuit:

  

Evaluating the Thevenin equivalent voltage at the base terminal:

  $\begin{array}{c}{V}_{TH}=\left(\frac{R_2}{R_1+R_2}\right)V_{CC}\\ =\left(\frac{220}{200+220}\right)\left(5\right)\\ =2.62\text{V}\end{array}$

Evaluating the Thevenin equivalent resistance at the base terminal:

  $\begin{array}{c}{R}_{TH}=R_1𑨀R_2\\ =\left(\frac{R_1{R}_2}{R_1+R_2}\right)\\ =\frac{200\times 220}{200+220}\text{kΩ}\\ =104.76\text{kΩ}\end{array}$

Drawing the DC equivalent circuit as shown below:

  

Evaluating the value of base current $I_B$ :

  $\begin{array}{c}{I}_B=\frac{V_{TH}-V_{BE\left(on\right)}}{R_{TH}+\left({\beta }_0+1\right)R_E}\\ =\frac{2.62-0.7}{\left(104.76\times {10}^3\right)+\left(100+1\right)\left(1\times {10}^3\right)}\\ =9.33\times {10}^{-6}\\ =9.33\text{μA}\end{array}$

Evaluating the value of collector current $I_C$ :

  $\begin{array}{c}{I}_C={\beta }_0{I}_B\\ =\left(100\right)\left(9.33\times {10}^{-6}\right)\\ =0.933\text{mA}\end{array}$

Evaluating the value of small signal parameters $g_m$ :

  $\begin{array}{c}{g}_m=\frac{I_C}{V_T}\\ =\frac{0.933\times {10}^{-6}}{26\times {10}^{-3}}\\ =35.88\times {10}^{-3}\\ =35.88\text{mA/V}\end{array}$

Evaluating the value of small signal parameters $r_{\pi }$ :

  $\begin{array}{c}{r}_{\pi }=\frac{{\beta }_o}{g_m}\\ =\frac{100}{35.88\times {10}^{-3}}\\ =2.78\text{k}\Omega \end{array}$

Drawing the small-signal equivalent circuit for the mid band current gain:

  

Applying the nodal analysis at the node $V_{\pi }$ :

  $\begin{array}{l}\frac{V_{\pi }}{r_{\pi }}+\frac{V_{\pi }}{R_{TH}}+\frac{V_{\pi }}{r_s}=I_s\\ V_{\pi }\left(\frac{1}{r_{\pi }}+\frac{1}{R_{TH}}+\frac{1}{r_s}\right)=I_s\\ V_{\pi }=\left(r_{\pi }𑨀R_{TH}𑨀r_s\right)I_s\end{array}$

Applying the current division rule at the output node:

  $\begin{array}{l}{I}_0=-\frac{R_C}{R_C+R_L}\left(g_m{V}_{\pi }\right)\\ I_0=-\frac{g_m{R}_C}{R_C+R_L}\left(r_{\pi }𑨀R_{TH}𑨀r_s\right)I_s\\ \frac{I_0}{I_s}=-\frac{g_m{R}_C}{R_C+R_L}\left(r_{\pi }𑨀R_{TH}𑨀r_s\right)\end{array}$

Evaluating the value of mid band current gain $A_i$ :

  $\begin{array}{c}{A}_i=-\frac{g_m{R}_C}{R_C+R_L}\left(r_{\pi }𑨀R_{TH}𑨀r_s\right)\\ =-\frac{\left(35.88\times {10}^{-3}\right)\left(2.2\times {10}^3\right)}{\left(2.2\times {10}^3\right)+\left(4.7\times {10}^3\right)}\left(\frac{1}{\frac{1}{104.76\times {10}^3}+\frac{1}{2.786\times {10}^3}+\frac{1}{100\times {10}^3}}\right)\\ =-\frac{\left(35.88\times {10}^{-3}\right)\left(2.2\times {10}^3\right)\left(2.642\times {10}^3\right)}{\left(6.9\times {10}^3\right)}\\ =-30.24\text{A/A}\end{array}$

Hence, the value of mid-band current gain $A_i$ is -30.24A/A.

To determine

The Miller capacitance C_M for the given values of the $C_{\mu }$ .

Answer to Problem 7.10EP

The Miller capacitances for the both cases are:

  $\begin{array}{c}\left(i\right)\text{For}{C}_{\mu }=0\\ C_M=0\\ \left(ii\right)\text{For }{C}_{\mu }=0.08\text{pF}\\ C_M=4.38\text{pF}\end{array}$

Explanation of Solution

Given:

The parameters for the given circuit:

  $\begin{array}{l}{R}_1=200\text{k}\Omega \\ R_2=220\text{k}\Omega \\ R_C=2.2\text{k}\Omega \\ R_L=4.7\text{k}\Omega \\ R_E=1\text{k}\Omega \\ r_S=100\text{k}\Omega \\ V_{CC}=5\text{V}\\ {\beta }_o=100\\ V_{BE\left(on\right)}=0.7\text{V}\\ V_A=\infty \\ C_{\pi }=1\text{pF}\end{array}$

The value of the capacitances are given as:

  $\begin{array}{l}\left(i\right)C_{\mu }=0\\ \left(ii\right)C_{\mu }=0.08\text{pF}\end{array}$

( i ) The value of capacitance $C_{\mu }$ :

  $C_{\mu }=0$

Evaluating the value of Miller capacitance for $C_{\mu }=0$ .

  $\begin{array}{c}{C}_M=C_{\mu }\left[1+g_m\left(R_C𑨀R_L\right)\right]\\ =\left(0\right)\left[1+g_m\left(R_C𑨀R_L\right)\right]\\ =0\end{array}$

Hence, the value of Miller capacitance for $C_{\mu }=0$ is $C_M=0$ .

(ii)

The value of capacitance:

  $C_{\mu }=0.08\text{pF}$

Evaluating the value of Miller capacitance for $C_{\mu }=0.08\text{pF}$ .

  $\begin{array}{c}{C}_M=C_{\mu }\left[1+g_m\left(R_C𑨀R_L\right)\right]\\ =\left(0.08\text{pF}\right)\left[1+\left(35.88\times {10}^{-3}\right)\frac{\left(2.2\times {10}^3\right)\left(4.7\times {10}^3\right)}{\left(2.2\times {10}^3\right)+\left(4.7\times {10}^3\right)}\right]\\ =\left(0.08\text{pF}\right)\left[1+\left(35.88\times {10}^{-3}\right)\left(1.49855\times {10}^3\right)\right]\\ =4.38\text{pF}\end{array}$

Hence, the value of the Miller capacitance for $C_{\mu }=0.08\text{pF}$ is $C_M=4.38\text{pF}$ .

To determine

The upper 3-dB frequency for the given values of the $C_{\mu }$ .

Answer to Problem 7.10EP

The 3-dB frequencies for the both cases are:

  $\begin{array}{c}\left(i\right)\text{For}{C}_{\mu }=0\\ f_{3dB}=60.2\text{MHz}\\ \left(ii\right)\text{For }{C}_{\mu }=0.08\text{pF}\\ f_{3dB}=11.2\text{MHz}\end{array}$

Explanation of Solution

Given:

The parameters for the given circuit:

  $\begin{array}{l}{R}_1=200\text{k}\Omega \\ R_2=220\text{k}\Omega \\ R_C=2.2\text{k}\Omega \\ R_L=4.7\text{k}\Omega \\ R_E=1\text{k}\Omega \\ r_S=100\text{k}\Omega \\ V_{CC}=5\text{V}\\ {\beta }_o=100\\ V_{BE\left(on\right)}=0.7\text{V}\\ V_A=\infty \\ C_{\pi }=1\text{pF}\end{array}$

The value of the capacitances are given as:

  $\begin{array}{l}\left(i\right)C_{\mu }=0\\ \left(ii\right)C_{\mu }=0.08\text{pF}\end{array}$

(i)

The value of Miller capacitance for $C_{\mu }=0$ :

Since, $C_M=0$ for this case.

Evaluating the upper $3\text{dB}$ frequency for $C_{\mu }=0$ :

  $\begin{array}{c}{f}_{3dB}=\frac{1}{2\pi \left(r_s𑨀R_{TH}𑨀r_{\pi }\right)\left(C_{\pi }+C_M\right)}\\ =\frac{1}{2\pi \left[\left(\frac{1}{\frac{1}{104.76\times {10}^3}+\frac{1}{2.786\times {10}^3}+\frac{1}{100\times {10}^3}}\right)\right]\left(1\times {10}^{-12}+0\right)}\\ =\frac{1}{2\pi \left(2.642\times {10}^3\right)\left(1\times {10}^{-12}\right)}\\ =60.2\text{MHz}\end{array}$

Hence, the upper $\text{3dB}$ frequency for $C_{\mu }=0$ is $f_{3dB}=60.2\text{MHz}$

(ii)

The value of Miller capacitance for $C_{\mu }=0.08pF$ is

Since, $C_M=4.38\text{pF}$ for this case.

Evaluating the upper $\text{3dB}$ frequency for $C_{\mu }=0.08pF$ .

  $\begin{array}{c}{f}_{3dB}=\frac{1}{2\pi \left(r_s𑨀R_{TH}𑨀r_{\pi }\right)\left(C_{\pi }+C_M\right)}\\ =\frac{1}{2\pi \left[\left(\frac{1}{\frac{1}{104.76\times {10}^3}+\frac{1}{2.786\times {10}^3}+\frac{1}{100\times {10}^3}}\right)\right]\left(1\times {10}^{-12}+4.38\times {10}^{-12}\right)}\\ =\frac{1}{2\pi \left(2.642\times {10}^3\right)\left(5.38\times {10}^{-12}\right)}\\ =11.2\text{MHz}\end{array}$

Hence, the upper $\text{3dB}$ frequency for $C_{\mu }=0.08\text{pF}$ is $f_{3dB}=11.2\text{MHz}$ .